Consider the addition of an electron to the following atoms from the third period. Rank the atoms in order from the most negativ
e to the least negative electron affinity values based on their electron configurations. Note: 1 = most negative electron affinity ; 3 = least negative electron affinity
1 answer:
The question is incomplete, the atoms in the third period referred to in the question are shown in the image attached.
Answer:
Cl>Si>Ar
Explanation:
Electron affinity refers to the energy released when an atom accepts electron (s) to form a negative ion. It is a periodic property that shows similar trends as electro negativity.
Considering the atoms in period three given in the question; chlorine will have the most negative value of electron affinity. This owes to the fact that halogens accept electrons easily to form negative ions. Silicon in group 14 shows much less negative value of electron affinity while argon, a noble gas, does not accept electrons at all and has an electron affinity of zero KJMol-1 by standard.
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Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
= 1 m³
= 10°C = 283 K
= 350 kPa
= 3 kg
= 35°C = 308 K
= 150 kPa
Now, lets apply the ideal gas equation;
=
R
=
R
/
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
= ( 3 × 0.287 × 308) / 150
= 265.188 / 150
= 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, =
/ R
= (350 × 1)/(0.287 × 283) = 350 / 81.221
= 4.309 kg
Total mass, =
+
= 4.309 + 3 = 7.309 kg
Total volume =
+
= 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
=
R
/
given that; final temperature = 20°C = 293 K
we substitute
= ( 7.309 × 0.287 × 293) / 2.77
= 614.6211119 / 2.77
= 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa